Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam.

### Solution:

Show me the final answerâ†“

Let us first draw a free body diagram showing all the forces like so:

We can now write a moment equation at point A to determine force B_y:

\circlearrowright^+ M_A=0;

600(4)+400\text{cos}\,(15^0)(12)-B_y(12)=0

B_y=586.37 N

600(4)+400\text{cos}\,(15^0)(12)-B_y(12)=0

B_y=586.37 N

Now, we can write an equilibrium equation for the x-axis forces.

\rightarrow^+\Sigma F_x=0;

A_x-400\text{sin}\,(15^0)=0

A_x=103.53 N

A_x-400\text{sin}\,(15^0)=0

A_x=103.53 N

Lastly, we can write an equilibrium equation for the y-axis forces.

\uparrow^+ \Sigma F_y=0;

A_y+B_y-600-400\text{cos}\,(15^0)=0

A_y+B_y-600-400\text{cos}\,(15^0)=0

###### (Remember we already found B_y=586.37 N)

A_y=400 N

Now that we know A_x and A_y, we can find the magnitude. Remember, magnitude is:

F_A=\sqrt{(A_x)^2+(A_y)^2}

F_A=\sqrt{(103.53)^2+(400)^2}

F_A=413.18 N

F_A=\sqrt{(103.53)^2+(400)^2}

F_A=413.18 N

### Final Answer:

F_A=413.18 N

F_B=B_y=586.37 N